हिंदी

1 / 2 ∫ 0 1 √ 1 − X 2 D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^{1/2} \frac{1}{\sqrt{1 - x^2}} dx\]

उत्तर

\[Let I = \int_0^\frac{1}{2} \frac{1}{\sqrt{1 - x^2}} d x . Then, \]
\[I = \left[ \sin^{- 1} x \right]_0^\frac{1}{2} \]
\[ \Rightarrow I = \sin^{- 1} \frac{1}{2} - \sin^{- 1} 0\]
\[ \Rightarrow I = \frac{\pi}{6} - 0\]
\[ \Rightarrow I = \frac{\pi}{6}\]

shaalaa.com
Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - Exercise 20.1 [पृष्ठ १६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.1 | Q 3 | पृष्ठ १६

संबंधित प्रश्न

\[\int\limits_0^{\pi/2} \cos^3 x\ dx\]

\[\int\limits_0^{\pi/2} \left( a^2 \cos^2 x + b^2 \sin^2 x \right) dx\]

\[\int\limits_0^1 \frac{1}{2 x^2 + x + 1} dx\]

\[\int\limits_1^4 \frac{x^2 + x}{\sqrt{2x + 1}} dx\]

\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]

 


\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]

\[\int_0^\frac{\pi}{2} \frac{\tan x}{1 + m^2 \tan^2 x}dx\]

\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]

\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]


\[\int\limits_0^1 \log\left( \frac{1}{x} - 1 \right) dx\]

 


Evaluate the following integral:

\[\int_{- 1}^1 \left| xcos\pi x \right|dx\]

 


\[\int_0^1 | x\sin \pi x | dx\]

If f(2a − x) = −f(x), prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 0 .\]

If f(x) is a continuous function defined on [−aa], then prove that 

\[\int\limits_{- a}^a f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( - x \right) \right\} dx\]

\[\int\limits_1^4 \left( x^2 - x \right) dx\]

\[\int\limits_0^2 \left( x^2 + x \right) dx\]

\[\int\limits_0^{\pi/4} \tan^2 x\ dx .\]

\[\int\limits_0^3 \frac{1}{x^2 + 9} dx .\]

\[\int\limits_0^{\pi/2} \log \left( \frac{3 + 5 \cos x}{3 + 5 \sin x} \right) dx .\]

 


\[\int\limits_a^b \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} dx .\]

If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.

 

 


If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]


\[\int\limits_1^e \log x\ dx =\]

The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .


The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is

 


\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]


\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]


Evaluate the following integrals :-

\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]


\[\int\limits_0^1 \left| 2x - 1 \right| dx\]


\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]


\[\int\limits_0^\pi \cos 2x \log \sin x dx\]


Using second fundamental theorem, evaluate the following:

`int_0^1 "e"^(2x)  "d"x`


Evaluate the following:

`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`


Evaluate the following using properties of definite integral:

`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x)  "d"x`


Evaluate the following:

`int_0^oo "e"^(-mx) x^6 "d"x`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.


`int x^9/(4x^2 + 1)^6  "d"x` is equal to ______.


The value of `int_2^3 x/(x^2 + 1)`dx is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×